Question: Solve for $x$. Enter the solutions from least to greatest. $(x + 1)^2 - 36 = 0$ $\text{lesser }x = $
Explanation: $\begin{aligned} (x + 1)^2 - 36&= 0 \\\\ (x+1)^2&=36 \\\\ \sqrt{(x+1)^2}&=\sqrt{36} \end{aligned}$ $\begin{aligned} x+1&=\pm6 \\\\ x&=\pm6-1 \\ \phantom{(x + 1)^2 - 36}& \\ x=-7&\text{ or }x=5 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -7 \\\\ \text{greater } x &= 5 \end{aligned}$